Školjka

%V0 Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^\circ$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ to $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and $$ \angle CHS - \angle CSB = 90^\circ, \quad \angle THC - \angle DTC = 90^\circ \text{.} $$ Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.